Question: Solve for $r$, $ \dfrac{4r + 6}{r - 2} = \dfrac{1}{3} $
Explanation: Multiply both sides of the equation by $r - 2$ $ 4r + 6 = \dfrac{r - 2}{3} $ Multiply both sides of the equation by $3$ $ 3(4r + 6) = r - 2 $ $12r + 18 = r - 2$ $11r + 18 = -2$ $11r = -20$ $r = -\dfrac{20}{11}$